V_{d d}=240ml=0,24 lít

n_NaOH=C_{M NaOH}.V=0,4.0,24=0,096(mol)

=> m_NaOH=n.M=0,096.40=3,84(g)\

n_KOH= C_{M KOH}.V=0,6.0,24=0,114 (mol)

=> m_KOH=n.M=0,114. 56=6,384(g)

Nhớ tick nhen ,chúc bạn học tốt

n_Fe=m/M=2,6/56=0,046(mol)

PT:

Fe + 2HCl -> FeCl₂ +H₂\(\uparrow\)

1..........2............1................1 (mol)

0,046->0,092->0,046-> 0,046 (mol)

V_H2=n.22,4=0,046.22,4=1,0304 lit

b) m_HCl=n.M=0,092.36,5=3,358(g)

c) m_{d d HCl}\(=\dfrac{m_{HCl}.100\%}{C\%}=\dfrac{3,358.100}{10}=33,58\left(g\right)\)

Dung dịch thu được sau phản ứng là FeCl₂

m_{d d sau phản ứng}=m_Fe + m_HCl - m_H2 =2,6+33,58-(0,046.2)=36,088 (g)

m_FeCl2=n.M=0,046.127=5,842(g)

=> \(C\%_{ddsauphanung}=\dfrac{m_{FeCl_2}.100\%}{m_{ddsauphanung}}=\dfrac{5,842.100}{36,088}\approx16,188\left(\%\right)\)

a) n_Zn=m/M=13/65=0,2(mol)

PT:

Zn + 2HCl -> ZnCl₂ +H₂ \(\uparrow\)

1..........2...........1............1 (mol)

0,2->0,4 -> 0,2 -> 0,2 (mol)

V_H2=n.22,4=0,2.22,4=4,48(lít)

b) m_HCl=n.M=0,4.36,5=14,6 (g)

=> \(C\%=\dfrac{m_{HCl}.100\%}{m_{ddHCl}}=\dfrac{14,6.100}{500}=2,92\left(\%\right)\)

c)m_ZnCl2=n.M=0,2.136=27,2(g)

n_H2=2,24/22,4=0,1(mol)

n_O2=1,68/22,4=0,075 (mol)

PT:

2H₂ + O₂ \(\underrightarrow{t^0}\) 2H₂O

2 .........1...............2 (mol)

0,1 -> 0,05 -> 0,1 (mol)

m_H2O=n.M=0,1.18=1,8(g)

Còn chất dư là O₂

Số mol O₂ dư 0,075-0,05=0,025(mol)

m_{O2 dư} =n.M=0,025.32=0,8(g)

n_Fe=m/M=11,2/56=0,2(mol)

V_HCl=100ml=0,1 (lit)

Pt:

Fe + 2HCl -> FeCl₂ +H₂\(\uparrow\)

1.........2............1..........1 (mol)

0,2 ->0,4 -> 0,2 ->0,2 (mol)

m_FeCl2=n.M=0,2.127=25,4(g)

C_{M HCl}=\(\dfrac{n}{V}=\dfrac{0,4}{0,1}=4\left(M\right)\)

c) V_H2=n.22,4=0,2.22,4=4,48(lít)

n_H2=V/22,4=5,6/22,4=0,25(mol)

PT:

Zn + 2HCl -> ZnCl₂ + H₂

1.........2............1............1 (mol)

0,25 < -0,5 <-0,25< - 0,25 (mol)

=> \(m_{Zn}=n.m=0,25.65=16,25\left(g\right)\)

n_Fe=m/M=2,8/56=0,05(mol)

a) PT:

Fe + 2HCl -> FeCl₂ + H₂\(\uparrow\)

1...........2............1................1 (mol)

0,05 ->0,1 - > 0,05 -> 0,05 (mol)

b) Khí thoát ra là H₂

V_H2=n.22,4=0,05.22,4=1,12(lít)

c) m_HCl=n.M=0,1.36,5=3,65(g)

=> m_{d d HCl} =\(\dfrac{m_{HCl}.100\%}{C\%}=\dfrac{3,65.100}{10}=36,5\left(g\right)\)

Nhớ tick nhé

n=50

tick nha

a) \(A=\left(\dfrac{3-x}{x+3}.\dfrac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}\right).\dfrac{x+3}{3x^2}\)

\(\Leftrightarrow A=\left(\dfrac{-\left(x-3\right)}{x+3}.\dfrac{\left(x+3\right)}{x-3}\right).\dfrac{x+3}{3x^2}\)

\(\Leftrightarrow A=\dfrac{-x-3}{3x^2}\)

b) Khi \(x=\dfrac{2}{3}\Rightarrow\) \(A=\dfrac{-\dfrac{2}{3}-3}{3.\left(\dfrac{2}{3}\right)^2}=\dfrac{-11}{4}\)

c) Để A < 0 thì

\(\dfrac{-x-3}{3x^2}< 0\)

=> -x -3 <0

<=> -x < 3

\(\Rightarrow x>3\)

a) Ta có: \(\left(a-b\right)^2\ge0\)

=>\(a^2+b^2-2ab\ge0\left(đpcm\right)\)

b) \(\left(a+b\right)^2\ge0\)

=> \(a^2+b^2+2ab\ge0\)

<=> \(a^2+b^2\ge-2ab\)

<=> \(\dfrac{a^2+b^2}{2}\ge ab\) (đpcm)

c) ta có: \(\left(a+1\right)^2=a^2+2a+1\)

\(a\left(a+2\right)=a^2+2a\)

Vậy từ 2 điều trên => \(a\left(a+2\right)< \left(a+1\right)^2\)

d) \(m^2+n^2+2\ge2\left(m+n\right)\) (*)

<=>m² - 2m +1 +n² - 2n +1 \(\ge0\)

<=> \(\left(m-1\right)^2+\left(n-1\right)^2\ge0\) (1)

(1) đúng => (*) đúng

d) Bạn ấy giải rồi ,mình không giải nữa